HDU 1866 A + B forever! 有坑，注意题意！！！

A + B forever!

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1080    Accepted Submission(s): 232

Problem Description

As always, A + B is the necessary problem of this warming-up contest. But the patterns and contents are different from the previous ones. Now I come up with a new “A + B” problem for you, the top coders of HDU.
As we say, the addition defined between two rectangles is the sum of their area . And you just have to tell me the ultimate area if there are a few rectangles.
Isn’t it a piece of cake for you? Come on! Capture the bright “accepted” for yourself.

 

Input

There come a lot of cases. In each case, there is only a string in one line. There are four integers, such as “(x1,y1,x2,y2)”, describing the coordinates of the rectangle, with two brackets distinguishing other rectangle(s) from the string. There lies a plus symbol between every two rectangles. Blanks separating the integers and the interpunctions are added into the strings arbitrarily.The length of the string doesn’t exceed 500.
0<=x1,x2<=1000,0<=y1,y2<=1000.

 

Output

For each case, you just need to print the area for this “A+B” problem. The results will not exceed the limit of the 32-signed integer.

 

Sample Input

(1,1,2,2)+(3,3,4,4)(1,1,3,3)+(2,2,4,4)+(5,5,6,6)

 

Sample Output

28

 

Author

Wang Ye

 

Source

2008杭电集训队选拔赛——热身赛

 

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(1,/1,3,;; 3) / + ,/ (2, 2,4,4) +  , ( %6  ,!!6,5,5)输出8

#include<stdio.h>#include<string.h>int m[1005][1005];int max(int a,int b) { return a>b?a:b; }int min(int a,int b) { return a<b?a:b; }int main(){char s[505],ch[505];int i,j,k,x1,y1,x2,y2,xmin,xmax,ymin,ymax,len,c,p;while(gets(s)){len=strlen(s);memset(m,0,sizeof(m)); memset(ch,0,sizeof(ch));c=k=p=0;for(i=0;i<len;i++){if(s[i]>='0'&&s[i]<='9'){ch[k++]=s[i];if(!(s[i+1]>='0'&&s[i+1]<='9')) p++,ch[k++]=' ';}if(p==4){ch[k++]='+';ch[k++]=' ';p=0;}}ch[k]='\0';len=strlen(ch);for(k=0;k<len;k++){if(k==0 || ch[k-1]=='+'){sscanf(ch+k,"%d %d %d %d",&x1,&y1,&x2,&y2);xmin=min(x1,x2); xmax=max(x1,x2);ymin=min(y1,y2); ymax=max(y1,y2);for(i=xmin;i<xmax;i++)for(j=ymin;j<ymax;j++)if(!m[i][j]) {c++; m[i][j]=1;}}}printf("%d\n",c);}return 0;}